Question

Use implicit differentiation to find an equation of the tangent line to the devil's curve defined by
y^2(y^2-9) = x^2(x^2-3^{2}+1) at the point (0,-3).

Answer 1

So the equation is

\[y^2(y^2-9)=x^2(x^2-3^2+1)\]

?

Answer 2

Yes!

Answer 3

OK first we are going to simplify it to make this easier for our sake. Distribute what we can into the parenthesis and also square the 3 to a 9 and we get.

\[y^4 - 9y^2 = x^4 - 9x^2 + x^2\]

You see how I got that?

Answer 4

Mhmm I do

Answer 5

Ok now here is the bugger. The derivative. Anytime we take a derivative of a term with y we need to tag it with dy/dx so lets do this a piece at a time

\[\frac{ d }{ dx } y^4 = 4y^3\frac{ dy }{ dx }\]
\[\frac{ d }{ dx } 9y^2 = 18y\frac{ dy }{ dx }\]
\[\frac{ d }{ dx }x^4 = 4x^3\]
\[\frac{ d }{ dx }9x^2 = 18x\]
\[\frac{ d }{ dx }x^2 = 2x\]

So then we have

\[4y^3\frac{ dy }{ dx } - 18y \frac{ dy }{ dx }=4x^3 -16x\]

This all make sense?

Answer 6

Yes!

Answer 7

now we solve to dy/dx by factoring it out and dividing by what is left

\[\frac{ dy }{ dx }(4y^3-18y) = 4x^3-16x\]
\[\frac{ dy }{ dx } = \frac{ 4x^3-16x }{ 4y^3-18x }\]

Answer 8

This is our derivative now so now we got a few more steps to find the tangent line

Answer 9

Okay! Thanks for your help so far :)

Answer 10

Ok the point tells us that x = 0 and y = -3 so we plug those into our derivative to find the slope value at that point but as we can see if you plug in a 0 for x you get 0 in the numerator meaning our slope value is 0 which makes our job easier

Set it up in slope intercept form

y=mx+b
y=0x+b
y+b

since y = -3

y=-3 is our tangent line

Answer 11

y=b not y+b sorry

Answer 12

So if we analyze this further it means that at (0,-3)

This graph must have a horizontal tangent line to it